# Replacing the $M$ coefficient in Schwarzschild metric with a distribution $M(r)$

Consider the Schwarzschild metric

$$ds^2 = -f(r)dt^2 + \frac{1}{f(r)}dr^2+r^2 d\Omega^2,$$

where $$f(r)=1-\frac{2M}{r}$$. I take it that $$M$$, although not really the mass of a black hole, is the coefficient that is closely associated with the mass. I'm curious as to what will happen if in the metric I directly replace $$M$$ with a distribution $$M(r)$$ and what happens to things like null and timelike geodesics, its horizons, etc.

My question is: Is it a valid starting point of inquiry to replace $$M$$ with a distribution $$M(r)$$? Or are there any physical rules that will make this problem ill-posed?

What will happen if in the metric I directly replace $$M$$ with a distribution $$M(r)$$?
1. Let us for simplicity work in units where the speed of light $$c=1$$ is equal to one, and assume that there is no cosmological constant $$\Lambda=0$$. A spherically symmetric vacuum solution to the EFE of the form $$ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,\tag{1}$$ and such that it asymtotically becomes Minkowski space $$-g_{tt}(r\!=\!\infty)~=~ 1~=~g_{rr}(r\!=\!\infty), \tag{2}$$ is then uniquely given by $$-g_{tt}(r)~=~ 1-\frac{R_S}{r} ~=~\frac{1}{g_{rr}(r)},\tag{3}$$ where $$R_S$$ is a length parameter, i.e. a constant, cf. Birkhoff's theorem and this Phys.SE post.
2. In particular, OP's function $$M(r)$$ can therefore not depend on $$r$$. For the interpretation of $$M$$ as the mass of the black hole, see this Phys.SE post.